Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

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Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Div2_419C {

	public static void main(String[] args) throws IOException {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter printer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		StringTokenizer inputData = new StringTokenizer(reader.readLine());
		int numR = Integer.parseInt(inputData.nextToken());
		int numC = Integer.parseInt(inputData.nextToken());

		int[][] tar = new int[numR + 1][numC + 1];
		for (int i = 1; i <= numR; i++) {
			inputData = new StringTokenizer(reader.readLine());
			for (int j = 1; j <= numC; j++) {
				tar[i][j] = Integer.parseInt(inputData.nextToken());
			}
		}

		int[] rowA = new int[numR + 1];
		int[] colA = new int[numC + 1];

		int[] rowAns = null;
		int[] colAns = null;

		int mCost = Integer.MAX_VALUE;
		rALoop:
		for (int fRA = 0; fRA <= tar[1][1]; fRA++) {
			int fCA = tar[1][1] - fRA;
			rowA[1] = fRA;
			colA[1] = fCA;

			int cost = tar[1][1];

			for (int i = 2; i <= numR; i++) {
				if (tar[i][1] < fCA) {
					continue rALoop;
				}
				rowA[i] = tar[i][1] - fCA;
				cost += rowA[i];
			}

			for (int i = 2; i <= numC; i++) {
				if (tar[1][i] < fRA) {
					continue rALoop;
				}
				colA[i] = tar[1][i] - fRA;
				cost += colA[i];
			}

			for (int i = 1; i <= numR; i++) {
				for (int j = 1; j <= numC; j++) {
					if (tar[i][j] != rowA[i] + colA[j]) {
						continue rALoop;
					}
				}
			}

			if (cost < mCost) {
				mCost = cost;
				rowAns = Arrays.copyOf(rowA, numR + 1);
				colAns = Arrays.copyOf(colA, numC + 1);
			}
		}

		if (mCost == Integer.MAX_VALUE) {
			printer.println(-1);
		} else {
			printer.println(mCost);
			for (int i = 1; i <= numR; i++) {
				for (int j = 0; j < rowAns[i]; j++) {
					printer.println("row " + i);
				}
			}
			
			for (int i = 1; i <= numC; i++) {
				for (int j = 0; j < colAns[i]; j++) {
					printer.println("col " + i);
				}
			}
		}
		printer.close();
	}

}